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October 1, 2014

October 1, 2014

Posted by **lindsay** on Sunday, March 25, 2007 at 5:13am.

1)A pocket of gas is discovered in a deep drilling operation. The gas has a temperature of 480c and is at a pressure of 12.8 atm. What volume of gas is required to provide 18.0 L of gas at the surface where the conditions are 22 C and 1.00 atm?

2)What is the volume of 45.0g of nitrogen monoxide, NO, at 20 C and a pressure of 740 mm Hg?

1) 480 C = 753 K; 22 C = 299 K

Let V2 be the volume at the surface (18.0 l) and V1 be the value at the low depth. You want to solve for V1.

P1 V1/T1 = P2 V2/T2, with temperatures in Kelvin.

(12.8 x V1)/753 = (1.00 x 19.0)/ 299

Solve for V1.

2) The molar mass is 16 + 12 = 28.00, so

45.0 g is n = 1.61 moles. That number of moles would occupy

22.4 x 1.61 = 36.06 liters at 1.00 atm and 273 K. At 20 C and 740 mm Hg(which corresponds to 0.974 atm), the volume has to be multiplied by (299/273)(1.000/0.973).

OR you could just use V = n R T/p, with R = 8.206*10^-2 l-atm/(mole K)

is this correct for number 1?

480c = 753k, 22c = 295k

p1 x v1/T1 = p2 x v2/T2

12.8 x v1/753 = 1.00 x 18.0/295

.016 = .061

.016 is V1

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